All the numbers from 100,000 to 111,111 contain a zero, so that is 11,111 numbers already. The question is then how many numbers between 1,111 and 2,000 contain at least one zero. There are 19 numbers every hundred digits from 1,200 onwards that meet this requirement (10 which have a 0 in the tens and ten which have a 0 in the units, not counting 00 twice) and thus we have 19x8 = 152 additional numbers. Add this to the amount from before, and 1 for the end value of 112,000: 11,111+152+1 = 11,264 numbers.
Greatest: 989949 Least: 100000
271 of the first 1000 natural numbers contain at least one digit 5. That is 27.1 % of them.
27: 077,177,277,377,477,577,677,877,977,717,727,737,747,757,767,777,787,797
9*10*10*10*10 - 8*9*9*9*9
To find the total number of seven-digit numbers that contain the number seven at least once, we can use the principle of complementary counting. There are a total of 9,999,999 seven-digit numbers in total. To find the number of seven-digit numbers that do not contain the number seven, we can count the number of choices for each digit (excluding seven), which is 9 choices for each digit. Therefore, there are 9^7 seven-digit numbers that do not contain the number seven. Subtracting this from the total number of seven-digit numbers gives us the number of seven-digit numbers that contain the number seven at least once.
Only one. 123456.
Greatest: 989949 Least: 100000
theres only 1 number which is 123456, the least digit is 1 and the greatest is 6
271 of the first 1000 natural numbers contain at least one digit 5. That is 27.1 % of them.
Equation
252
27: 077,177,277,377,477,577,677,877,977,717,727,737,747,757,767,777,787,797
9*10*10*10*10 - 8*9*9*9*9
i think alot like 100 at least
46000
252
To find the total number of seven-digit numbers that contain the number seven at least once, we can use the principle of complementary counting. There are a total of 9,999,999 seven-digit numbers in total. To find the number of seven-digit numbers that do not contain the number seven, we can count the number of choices for each digit (excluding seven), which is 9 choices for each digit. Therefore, there are 9^7 seven-digit numbers that do not contain the number seven. Subtracting this from the total number of seven-digit numbers gives us the number of seven-digit numbers that contain the number seven at least once.